Question

Money magazine reported that 30% of 1013 adults telephoned at random could not correctly define any of the four main types of life insurance. The article cautioned that a "3.1% margin of error" was built into that figure a. Most polls report a "margin of error" along with their results. While it is not typically stated, this margin of error is almost always based on a 95% confidence interval for the true population proportion p for each question. Use our traditional formula ptza/2 to find a 95% confidence interval based on this data. Does your margin of error agree with the 3.1% reported by the magazine? b. Show that the largest possible value of occurs when p 0.5 (assuming n is fixed) Most polls ask multiple questions. As a result, in reality, this margin of error will be different for each different question asked in the poll, because the value of p used to compute the margin of error will be different for each different question in the poll. However, most polls report only a single margin of error for all questions on the poll. They do this by replacing p with 0.5 in the computation of the margin of error c. Recompute the margin of error for the data above using p-0.5 instead of the actual value of p Does your result now agree with the value of 3.1% reported by the magazine? How does this value compare to the value you found in part (a)? d. We designed our confidence interval formula so that 0.95. What will always happen to this 1 1 probability if p is replaced with 0.5 in the margin of error computation? What impact will this have on the actual level of confidence? e. Explain as precisely as possible in a short paragraph what the 3.1% reported by the magazine means

Answer 1

From the sample we have

n=1013 is the sample size

is the sample percentage of adults who could not define any of the 4 types of life insurance.

p 0.30

a) The 95% confidence interval indicates that the level of significance

a-1-95/1000.05

The critical value is

P(Z > z/2)-a/2= 0.05/2 0.02 P(Z < a/2 )-1-0.025-0.975

Using the standard normal tables we get that for z=1.96 P(Z<1.96) = 0.975.

Hence

2i 1.96

The 95% confidence interval for true population proportion is

0.301-0.30) 0.30 ± 1.96 1013 0.30 ± 0.0282 0.272,0.328]

ans: 95% confidence interval for the true population proportion is [27.2%, 32.8%]

The margin of error is

p(1- 0.30(1 0.30) 1.96 1013 -0,028

In terms of percentages the margin of error is 2.82%

Ans: The margin of error that we calculated does not agree with the 3.1% reported by the magazine.

b) We want to find the value of $\begin{align*} \hat{p} \end{align*}$ which will maximize the value of .

p 1-p)

The first order condition is to equate the first derivative of
p 1-p)
w.r.t $\begin{align*} \hat{p} \end{align*}$ to zero

d p(1 - p) dp x(1-2) = 0 The first term 2ympI- cannot be=0 for finite n and 0 < þ < 1 hence (1-2) = 0 P0.5

ans: the largest possible value of
p 1-p)
occurs for $\begin{align*} \hat{p}=0.5 \end{align*}$

c) The margin of error with $\begin{align*} \hat{p}=0.5 \end{align*}$ is

(1-p) 0.50(1 0.50) = 1.96 1013 0.0308

That is in terms of percentages, the margin of error is 3.08% or can be rounded to 3.1%

ans: With $\begin{align*} \hat{p}=0.5 \end{align*}$ , the margin of error agree with the value of 3.1% reported by the magazine. This value is larger than the value found in part a).

d) The formula for the confidence interval is

$\begin{align*}&P\left(\hat{p}-z_{\alpha/2}\sqrt\frac{\hat{p}(1-\hat{p})}{n}<p<\hat{p}+z_{\alpha/2}\sqrt\frac{\hat{p}(1-\hat{p})}{n} \right )\\ &\text{if we subtract }\hat{p}\text{ and divide by} \sqrt\frac{\hat{p}(1-\hat{p})}{n}\\ &\text{we get}\\ &P\left(-z_{\alpha/2}<\frac{p-\hat{p}}{\sqrt\frac{\hat{p}(1-\hat{p})}{n}}<+z_{\alpha/2}\right )\\ \end{align*}$

Due to the central limit theorem, we can say that the term

$\begin{align*}\frac{p-\hat{p}}{\sqrt\frac{\hat{p}(1-\hat{p})}{n}} \end{align*}$ has a standard normal distribution N(0,1)

and the value of for a 95% confidence interval.

2i 1.96

Hence

$\begin{align*}P\left(-1.96<\frac{p-\hat{p}}{\sqrt\frac{\hat{p}(1-\hat{p})}{n}}<+1.96\right )=P\left(-1.96<Z<+1.96\right )=0.95\\ \end{align*}$

irrespective of the value of $\begin{align*} \hat{p} \end{align*}$ .

When $\begin{align*} \hat{p} \end{align*}$ is replaced by 0.5, the margin of error increases, but the level of confidence remains the same at 95%

e) 3.1% of margin of error means that there is a 95% chance that the true population proportion of the percentage of adults who cannot correctly define any of the four main types of life insurance lies within +-3.1% of 30% (that is between 30-3.1=26.9% and 30+3.1=33.1%)